Question

If the average salary in a company is $52,500 and the standard deviation is $7,800, what...

If the average salary in a company is $52,500 and the standard deviation is $7,800, what percentage of employees earn more than $60,300?

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Answer #1

Desired probability

P(X > 60300) = P(z > 6030052500 7800

From standard Normal Table we get the value of P(z > 1) = 0

So, 0% of people earn more than 60,300$ assuming the distribution to be normal.

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